Integrated Scheduling of Production and Distribution Operations

MANAGEMENT SCIENCE

informs

Vol. 51, No. 4, April 2005, pp. 614–628 issn 0025-1909 eissn 1526-5501 05 5104 0614

®

doi 10.1287/mnsc.1040.0325 © 2005 INFORMS

Integrated Scheduling of Production and Distribution Operations Zhi-Long Chen

Department of Decision and Information Technologies, Robert H. Smith School of Business, University of Maryland, College Park, Maryland 20742-1815, [email protected]

George L. Vairaktarakis

Department of Operations, Weatherhead School of Management, Case Western Reserve University, Cleveland, Ohio 44106-7235, [email protected]

M

otivated by applications in the computer and food catering service industries, we study an integrated scheduling model of production and distribution operations. In this model, a set of jobs (i.e., customer orders) are first processed in a processing facility (e.g., manufacturing plant or service center) and then delivered to the customers directly without intermediate inventory. The problem is to find a joint schedule of production and distribution such that an objective function that takes into account both customer service level and total distribution cost is optimized. Customer service level is measured by a function of the times when the jobs are delivered to the customers. The distribution cost of a delivery shipment consists of a fixed charge and a variable cost proportional to the total distance of the route taken by the shipment. We study two classes of problems under this integrated scheduling model. In the first class of problems, customer service is measured by the average time when the jobs are delivered to the customers; in the second class, customer service is measured by the maximum time when the jobs are delivered to the customers. Two machine configurations in the processing facility—single machine and parallel machine—are considered. For each of the problems studied, we provide an efficient exact algorithm, or a proof of intractability accompanied by a heuristic algorithm with worst-case and asymptotic performance analysis. Computational experiments demonstrate that the heuristics developed are capable of generating near-optimal solutions. We also investigate the possible benefit of using the proposed integrated model relative to a sequential model where production and distribution operations are scheduled sequentially and separately. Computational tests show that in many cases a significant benefit can be achieved by integration. Key words: integrated scheduling; production and distribution operations; dynamic programming; heuristic; worst-case performance analysis; asymptotic performance analysis; value of integration History: Accepted by Dorit Hochbaum, optimization and modeling; received August 1, 2003. This paper was with the authors 6 12 months for 4 revisions.

1.

Introduction

only after orders are known; hence no inventory is involved and completed orders are delivered within a few hours. In both systems, production operations (i.e., assembly and packaging of computers, food preparation) and distribution operations (i.e., delivery of completed orders to customers) are linked together directly without any intermediate step. In such make-to-order applications, cost and customer service are the major concerns of the decision maker. Very little inventory of finished products exists at any point of time, making inventory cost negligible. The total cost of the system is mainly contributed by the production and distribution operations. The cost of a delivery shipment usually consists of a fixed charge and a variable cost proportional to the total distance of the route taken. Thus, the number of shipments used and the specific routes taken by these shipments determine the total distribution cost. On the contrary, the total production cost for a given

Many companies produce and deliver to customers directly without holding intermediate inventories. This is particularly true in the computer and food catering service industries, where direct orders are an important part of business. In a typical computer direct order system, there are hundreds of configurations available for a customer to choose from when ordering. It is thus impractical to make to stock based on forecasts of each configuration. Consequently, assembly and packaging operations can take place only after customer orders come in. On the other hand, because of fierce competition in the computer industry, completed orders (i.e., assembled and packaged computers) are normally delivered to customers within a short lead time (e.g., two to three business days). Similar characteristics can be seen in a typical direct order system of food catering services. To keep food fresh, food preparation can take place 614

Chen and Vairaktarakis: Integrated Scheduling of Production and Distribution Operations Management Science 51(4), pp. 614–628, © 2005 INFORMS

set of orders is normally fixed and independent of the production schedule used. So one may just treat total distribution cost as the only cost factor involved in the system. In terms of customer service level, generally the shorter the lead times, the higher the service level. To achieve shorter lead times, more delivery shipments may have to be used, which will result in higher distribution cost. Therefore, optimizing the trade-off between the distribution cost and the customer service level is often a major goal of the decision maker of such systems. The close linkage between production and distribution necessitates coordinating production and distribution operations at the level of detailed scheduling. Motivated by the above-described applications in the computer and food catering service industries, in this paper we study an integrated scheduling model of production and distribution operations. A set of jobs (i.e., customer orders) are first processed in a processing facility (e.g., manufacturing plant or service center) and then delivered to customers directly without intermediate inventory. The problem is to find a joint schedule of production and distribution such that an objective function that takes into account both customer service level and total distribution cost is optimized. Customer service level is measured by a function of the times when the completed jobs are delivered to the customers. A production schedule specifies the jobs assigned to each machine and the processing sequence of the jobs on each machine in the processing facility. Finished jobs are delivered to customers by homogeneous vehicles. We assume that distribution operations are carried out by third-party carriers, and hence an unlimited number of vehicles are available. As a result, each vehicle is used at most once, and each shipment is delivered by a different vehicle. However, due to limited vehicle capacity, each shipment can only carry up to a prespecified number of jobs. A distribution schedule specifies how many shipments to use, the jobs composing each shipment, the departure time of each shipment from the processing facility, and the route followed by each shipment. The cost of each delivery shipment consists of a fixed charge and a variable cost proportional to the total distance of the associated route. This model integrates production scheduling for job processing in the processing facility with distribution scheduling and routing for delivery of completed jobs from the processing facility to the customers. Although production-scheduling problems and vehicle-routing problems have both been extensively studied in the literature (e.g., Pinedo 2002, Bramel and Simchi-Levi 1997), little work has been done on the integration of production scheduling and vehicle routing. Existing models are either special cases of our model or have a different structure. They either consider

615

customer service as their objective, without accounting for transportation costs, or assume that jobs can be delivered to customers instantaneously without any transportation delay. Furthermore, very few of them consider routing decisions. We briefly review these models next. Potts (1980), Hall and Shmoys (1992), and Woeginger (1994, 1998) study a model in which a set of jobs are first processed in a processing facility and then delivered to customers, and the objective is to optimize customer service. There are sufficient delivery vehicles available such that each job is delivered by a separate shipment to its customer immediately after it is processed. Therefore, production scheduling is the only decision to be made. Lee and Chen (2001) consider a model where all the jobs belong to a single customer and the number of delivery vehicles is limited such that a vehicle may have to deliver multiple shipments. Hall et al. (2001) investigate a model where there is a fixed set of delivery dates at which completed jobs can be delivered. None of the above works consider transportation costs as part of the objective, and no routing decisions are involved in these models. Li et al. (2005) and Geismar et al. (2003) study problems involving routing decisions, but again without transportation costs. In the problem studied by Li et al., only a single delivery vehicle is available, all the jobs belong to one or two customers, and the objective is to deliver all the jobs in the minimum possible time. The problem considered by Geismar et al. has a similar objective but has a general number of customers and a constraint that a job has to be delivered within a prespecified amount of time after it is processed. Herrmann and Lee (1993), Chen (1996), Cheng et al. (1996), and Hall and Potts (2003) study a different set of models that consider both customer service and transportation cost as part of the objective but assume that job delivery can be done instantaneously without any transportation time. In these models the transportation cost of each shipment is fixed, and no vehicle-routing decision is involved. Another set of models somewhat related to ours are found in joint production, inventory, and transportation problems in the supply chain management literature (e.g., Cohen and Lee 1988, Chandra and Fisher 1994, Hahm and Yano 1995, Fumero and Vercellis 1999, Sarmiento and Nagi 1999). As pointed out by Chen (2004), most existing production-inventorydistribution models study strategic or tactic levels of decisions, and few have addressed integrated decisions at the detailed job-scheduling level. In all these models production and distribution are indirectly linked through inventory, and inventory cost is a significant portion of the total cost. None of these models is applicable to the scheduling problems encountered in direct order systems that motivate our study. From

616


a broader perspective, recent reductions in inventory levels (Feld 2000, Rajagopalan and Malhotra 2001) have created a closer linkage between production and distribution in practice. Clearly, more academic research is needed to model direct productiondistribution interactions and develop problem-solving techniques that can be used in practice. The rest of this paper is organized as follows. In §2, we define the problems and provide optimality properties. We study problems with two different service measures in §§3 and 4, respectively. For each problem we provide an efficient exact algorithm or a proof of intractability and a heuristic with worstcase and asymptotic performance analysis. In §5, we investigate the benefit of integrating production and distribution operations as compared to a sequential model, where production and distribution operations are scheduled sequentially and separately. We conclude the paper in §6.

2.

Model and Preliminary Results

In this section, we describe our model and provide some optimality properties to be used later. There are n jobs N = 1 2 n from k customers K = 1 2 k, which are located at different locations in an underlying transportation network. Let Ni ⊂ N be the subset of the jobs destined for customer i, and ni = Ni for i ∈ K, where N = N1 ∪ · · · ∪ Nk and n = n1 + · · · + nk . Our model consists of a production part in a processing facility where all the jobs are processed and a distribution part that accounts for delivery of completed jobs to their respective customers. In the production part, we consider two of the most widely studied machine configurations: single machine and parallel machine. In the single-machine case, only one machine is available for processing all jobs, while in the parallel-machine case, m m ≥ 2 identical parallel machines are available and each job is processed by one of the machines. In both cases, each job j ∈ N has a given processing requirement of pj time units, which has to be executed on a machine without interruption. All jobs and machines are available at time 0. Given a production schedule, we use Cj to denote the completion time of job j ∈ N . Clearly, job j can only be picked up by a vehicle for delivery at or after time Cj . In the distribution part, there are enough homogeneous vehicles available so that each vehicle will be used at most once and each delivery shipment will be transported by a dedicated vehicle. All the vehicles are stationed at the processing facility at time 0 and must go back to the facility once they complete a shipment. Each vehicle can carry up to c jobs in one shipment. The transportation cost incurred by each shipment consists of a fixed charge f and

a variable cost dependent on the particular route taken by the vehicle. We use c0i , cij , and ci0 , respectively, to denote the variable cost for traveling from the processing facility to customer i ∈ K, from customer i ∈ K to customer j ∈ K, and from customer i ∈ K to the processing facility. The corresponding travel times are denoted as t0i tij , and ti0 , respectively. A delivery vehicle can depart from the processing facility only when all the jobs to be delivered have completed processing. We use Dj to denote the time when job j ∈ N is delivered to its customer. The problem is to find a joint production and distribution schedule such that a function measuring customer service level and total distribution cost, denoted as SD1 Dn + 1 − T , is minimized. Here, SD1 Dn measures customer service level and is a nondecreasing function of the delivery times D1 Dn of jobs, T denotes the total distribution cost, and is a given constant 0 ≤ ≤ 1 representing the decision maker’s relative preference on customer service level and total distribution cost. We consider two particular forms of the customer service function SD1 Dn : (i) maximum delivery time, i.e., SD1 Dn = Dmax = maxD1 Dn , and (ii) mean delivery time, i.e., SD1 Dn = Dmean = D1 + · · · + Dn /n, which represent the maximum and average order fulfillment lead time, respectively. Note that in the production-scheduling literature, two similar functions—makespan (i.e., Cmax = maxC1 Cn and mean completion time (i.e., Cmean = C1 + · · · + Cn /n — are widely used to measure customer service. In traditional scheduling models it is implicitly assumed that once a job is completed it is delivered to its customer instantaneously; hence, Cj is treated as the time job j is received by its customer. However, in our model, there are nonnegligible delivery times involved; hence, the functions Dmax and Dmean more accurately measure customer service. We consider a number of variations of our model. For ease of presentation, we borrow the three-field notation abc widely used in the production-scheduling literature (e.g., Pinedo 2002) to precisely represent our problems. Field a specifies the machine configuration in the processing facility. For singlemachine and m-parallel-machine cases, a = 1 and a = Pm , respectively. Field b ∈ 1 k represents the number of customers involved in the problem. If b = 1, there is only one customer (i.e., k = 1). If b = k, there are k ≥ 2 customers. The last field c is the objective function to be minimized. For example, Pm 1 Dmean + 1 − T represents the problem with one customer, m parallel machines, and the objective of minimizing Dmean + 1 − T . We study the following eight variations of our integrated scheduling model,


four with the customer service measure Dmean and four with measure Dmax : (P1) 11Dmean + 1 − T . (P2) 1kDmean + 1 − T . (P3) Pm 1Dmean + 1 − T . (P4) Pm kDmean + 1 − T . (P5) 11Dmax + 1 − T . (P6) 1kDmax + 1 − T . (P7) Pm 1Dmax + 1 − T . (P8) Pm kDmax + 1 − T . We present some straightforward optimality properties for all the problems. Proofs are omitted. Lemma 1. For all problems there exists an optimal schedule such that in the production part of the schedule there is no idle time between the jobs processed on each machine in the processing facility. A set of jobs is in SPT order (shortest-processingtime-first order) if the jobs are sequenced in a nondecreasing order of their processing times, and jobs with equal processing times are sequenced in the order of their indices. According to this definition, there is exactly one SPT order for a given set of jobs. Lemma 2. For all problems there exists an optimal schedule in which the jobs that belong to the same customer and are processed on the same machine in the processing facility are sequenced in SPT order. Lemma 3. For all problems there exists an optimal schedule such that— (i) the departure time of each shipment is the completion time of the last job included in the shipment. (ii) jobs that are processed on the same machine and delivered in the same shipment are processed consecutively on that machine.

3.

Solving Problems P1, P2, P3, and P4

In this section, we consider the first four problems P1, P2, P3, and P4, where customer service is measured by the mean delivery time Dmean . In §3.1, we propose efficient algorithms for finding optimal solutions or demonstrate that they are unlikely to exist. In §3.2, we propose a heuristic for NP-hard problems P3 and P4 with 0 < < 1, analyze its worst-case and asymptotic performance, and conduct a computational experiment.

3.1. Optimal Algorithms Solving these problems under two extreme cases of , = 0 or = 1 is generally different from the case where 0 < < 1, so we treat the problems with = 0 or = 1 separately. We first consider problem P1 with 0 < < 1. Because k = 1 and m = 1 in P1, by Lemma 2, there exists an optimal schedule where all the jobs are processed in SPT order on the machine. The problem is thus reduced to finding an optimal distribution

617

schedule given that the jobs are processed in SPT order on the machine. By Lemma 3, the problem is equivalent to partitioning the SPT order of the jobs into blocks of consecutive jobs such that each block is delivered by a shipment that departs at the time when the last job in it is completed. Because there is only one customer, no routing is involved and each shipment takes the same route at the same transportation cost f + c01 + c10 . Without loss of generality, we assume that the jobs are indexed in SPT order so that p1 ≤ · · · ≤ pn , and hence the completion time of each job j ∈ N is Cj = p1 + · · · + pj . The following dynamic programming algorithm finds an optimal schedule for P1 with 0 < < 1. Algorithm DP1 Define value function V j = minimum total cost contribution of a schedule for the first j jobs 1 j. Initial condition: V 0 = 0. Recursive relation: For j = 1 n, V j = min V j − h + hCj + t01 /n + 1 − · f + c01 + c10 h = 1 minc j Optimal solution value: V n . Theorem 1. Algorithm DP1 finds an optimal schedule for problem P1 with 0 < < 1 in On log n + nc time. Proof. By Lemma 3, in a schedule for the first j jobs 1 j, where the production schedule follows SPT order, the departure time of the last shipment is always Cj , regardless of the size of the last shipment. This implies that the delivery time of all the jobs in the last shipment is Cj + t01 . In the recursive relation, the value function is computed by trying every possible size of the last shipment. Given that the size of the last shipment is h (i.e., the last shipment delivers jobs j − h + 1 j − h + 2 j), the total cost contributed by the last shipment is hCj + t01 /n + 1 − f + c01 + c10 , where hCj + t01 /n is the contribution to the mean delivery time and f + c01 + c10 is the transportation cost. This proves the correctness of the recursive relation and hence the optimality of the algorithm. There are n states in this DP, and it takes no more than Oc time to calculate the value for each state. Sorting the jobs in SPT order takes On log n time. Therefore, the overall time complexity of DP1 is bounded by On log n + nc . Next, we consider problem P2 with 0 < < 1. By Lemma 2, there exists an optimal schedule for P2 where the jobs from each customer are in SPT order on the machine. However, on each machine all the jobs together do not necessarily form an SPT order. So, unlike P1, for which we only need to find an optimal distribution schedule given the SPT production schedule, in P2 we cannot treat production and distribution separately and have to optimize them jointly.

Chen and Vairaktarakis: Integrated Scheduling of Production and Distribution Operations

618

Management Science 51(4), pp. 614–628, © 2005 INFORMS

For ease of presentation, we reindex the jobs in Ni from each customer i ∈ K as i1 ini such that pi1 ≤ · · · ≤ pini and hence their SPT order is i1 ini . Define set Qih = i1 ih and Riuv = Qiv \ Qi v−u = iv − u + 1 iv − u + 2 iv , and define parameter Pih = hl=1 pil for h = 1 ni , v = u ni , u = 1 c, and i ∈ K. Algorithm DP2 Define value function V j1 jk = minimum total cost contribution of a schedule for the j1 + · · · + jk jobs from ki=1 Qi ji . In addition, define ti $ X = time needed to reach customer i from the processing facility in a given route $ covering all the customers associated with the jobs in the set X. g$ q1 qk = total variable transportation cost of a given route $, which delivers qu jobs to customer u for u = 1 k. (Hence, route $ visits all the customers u with qu ≥ 1.) Initial condition: V 0 0 = 0. Recursive relation: For ji = 0 1 ni ; i = 1 2 k. V j1 jk = min V j1 − q1 jk − qk k q P + ti $ Rl ql jl + n i∈K i l∈K ljl l=1

Theorem 3. Problems P3 and P4 with 0 < < 1 are NP-hard even if the number of customers is fixed.

+ 1 − f + g$ q1 qk all possible routes $( qi ≤ ji for

possible routes (at most k!) for visiting all the customers u with qu ≥ 1. Given a route, it takes no more than Ok time to calculate the total variable transportation cost g$ q1 qk . Therefore, the overall time complexity of the algorithm is bounded by Onk c k k!k ≤ Oc k nk kk−1 because k!/kk ≤ 2/k2 when k ≥ 2. Theorem 2 means that when the number of customers k is fixed, Algorithm DP2 solves P2 with 0 < < 1 in polynomial time. When the number of customers k is arbitrary, then problem P2 with 0 < < 1 is strongly NP-hard because it contains the traveling salesman problem (TSP), which is strongly NP-hard (Karp 1972), as a special case with pj = 0 for j ∈ N , tij = 0 for i j ∈ 0 k, c = n, and f sufficiently large such that all the jobs will be delivered in one shipment. Problems P3 and P4 with 0 < < 1 can be viewed as an extension of the classical parallel-machine scheduling problem of minimizing mean completion time Pm Cmean with the added distribution part. It is known (Conway et al. 1967) that for problem Pm Cmean , sequencing jobs in SPT order and assigning each job to the first available machine is optimal. However, problems P3 and P4 with 0 < < 1 are NP-hard even if the number of customers is fixed. This is proved in the following theorem.

i = 1 k( and 1 ≤ q1 + · · · + qk ≤ c Optimal solution value: V n1 nk . Theorem 2. Algorithm DP2 finds an optimal schedule for problem P2 with 0 < < 1 in Oc k nk kk−1 time. Proof. Suppose that only the first ji jobs of customer i (i.e., the jobs in Qi ji for i ∈ K are scheduled in a partial solution of the problem. If we know that the last shipment consists of qi jobs from each customer i, then by Lemmas 2 and 3, these qi jobs must be the jobs in Ri qi ji . The dynamic program compares all possible routes of the current last ship ment that contains the jobs in kl=1 Rl ql jl . In the recur sive relation, time of the l∈K Pljl is the departure k current last shipment and l∈K Pljl + ti $ l=1 Rl ql jl is the delivery time of the jobs in Ri qi ji for i ∈ K. This proves the optimality of DP2. There are at most nk states j1 jk and at most c k combinations of q1 qk satisfying 1 ≤ q1 + · · · + qk ≤ c. To calculate the value function V j1 jk for each state j1 jk , the algorithm enumerates all possible combinations of (q1 qk with 1 ≤ q1 + · · · + qk ≤ c. For each combination of q1 qk , it enumerates all

Proof. We prove the result for P3 with 0 < < 1 by a reduction from the known ordinarily NP-hard partition problem (Garey and Johnson 1979). Because P4 is more general than P3, P4 with 0 < < 1 is also NP-hard. Partition. Given h elements with integer sizes x1 xh , where hi=1 xi = 2X, does there exist a partition H1 , H 2 of the index set H = 1 h such that x = i∈H1 i i∈H2 xi = X? Consider an instance of the recognition version of problem P3 defined by Number of jobs n = h, job set N = H = 1 2 h Number of machines m = 2 Job processing times pj = xj for j ∈ N Travel time t01 = t10 = 0 Travel cost c01 = c10 = 0 Vehicle capacity c = h Fixed cost of a shipment f = 4X/1 − Threshold cost Y = X + 41 − X/1 − We show that there is a schedule to this instance of P3 if and only if there is a solution to the partition. If Part. Given a partition H1 and H2 of H , we construct a schedule for the instance of P3 as follows: In the production part, schedule jobs from Hi on machine i in an arbitrary order for i = 1 2; in the distribution part, use only one shipment to deliver all the jobs. The shipment departs at time X. It is easy


619


to see that this is a feasible schedule and its total cost is Y . Only If Part. Given a schedule for the instance of P3 with total cost no more than Y , we can conclude that there is only one shipment in this schedule, because otherwise the distribution cost would be no less than 2f and the total cost would be no less than 21− f > aX + 1 − f = Y . Now suppose that H1 and H2 are the sets of jobs processed on machines 1 and 2, respectively, in the processing facility. Then, the departure time of the shipment is xi xi ≥ X C = max i∈H1

i∈H2

Because the total cost is no more than Y , the departure time of the shipment must be C = X. This means that H1 and H2 must be a partition of H . Finally, we consider the problems with = 0 or = 1. If = 1, then the objective of these problems is to minimize Dmean and the transportation cost is not a part of the objective function. Then, each job is delivered separately and immediately after it is completed processing. Therefore, Cj + t0i /n = Cj /n + t0i /n Dmean = i∈K j∈Ni

j∈N

i∈K j∈Ni

This means that minimizing Dmean is equivalent to minimizing the average completion time Cmean of the jobs in the processing facility. We have the following result (see Conway et al. 1967). Remark 1. For problems P1, P2, P3, and P4 with = 1, it is optimal to first order the jobs in SPT sequence and then assign each job based on this sequence to the machine (for P1 and P2) or to the first available machine (for P3 and P4). If = 0, then the objective is to minimize the total transportation cost T irrespective of the production schedule. The following simple algorithm solves problems P1 and P3 with = 0: Algorithm A1 Step 1. Schedule jobs on the machine (for P1) or machines (for P3) in the processing facility in an arbitrary order. Reindex the jobs as -1. -n. such that C-1. ≤ · · · ≤ C-n. . Step 2. Let L = n/c. Form L shipments such that shipment j for j = 1 L − 1, delivers c jobs -j − 1 · c + 1. -j − 1 c + 2. -jc. and departs at time C-jc. , and shipment L delivers the last n − L − 1 c jobs -L − 1 c + 1. -L − 1 c + 2. -n. and departs at time C-n. . Algorithm A1 requires On time. The total transportation cost of the resulting schedule is Lf + t01 + t10 , which is the minimum possible for P1 and P3 with = 0. Hence, we have: Remark 2. Algorithm A1 solves P1 and P3 with = 0 optimally in On time.

For problems P2 and P4 with = 0, we can apply Algorithm DP2 after setting = 0 and defining the value function V j1 jk as the minimum total transportation cost of a schedule containing ji jobs from Ni for i = 1 k. As in Theorem 2, it can be proved that Algorithm DP2 finds an optimal schedule for problems P2 and P4 with = 0 in Oc k nk kk−1 time. Hence, when the number of customers k is fixed, problems P2 and P4 with = 0 are solvable by DP2 in polynomial time. When k is arbitrary, these problems become strongly NP-hard because they contain the strongly NP-hard TSP as a special case with c = n and f sufficiently large such that all the jobs will be delivered in one shipment. 3.2. Heuristic and Performance Analysis In this section, we give a heuristic for the NP-hard problems P3 and P4 with 0 < < 1. The heuristic is based on the following idea: Given problem P3 or P4, we first define a corresponding auxiliary problem, which is solved by a modified version of Algorithm DP2 described in §3.1. The optimal solution of the auxiliary problem is then transformed to a feasible solution for P3 or P4. Given P3 or P4, define an auxiliary problem AUX as follows: Partition the job set N into subsets S1 S2 such that the number of jobs in each subset Sq is no more than c. Deliver each subset of jobs Sq in one shipment, with the departure time dq required to be the maximum between (i) the maximum processing time of the jobs delivered in the shipments prior to and including Sq (i.e., maxpj j ∈ S1 ∪ · · · ∪ Sq ), and (ii) the sum of the processing times of the jobs delivered in the shipments prior to and including Sq divided by m (i.e., j∈S1 ∪···∪Sq pj /m). The problem is to partition N into shipments S1 S2 so as to minimize the objective function Dmean + 1 − T subject to the above-stated constraints. Evidently, AUX differs from problems P1 and P2 only in the way we calculate the departure time of a shipment. Problem AUX uses the maximum of (i) and (ii) while P1 and P2 use the completion time of the job completed last in a shipment. We will show that when the recursive relation in Algorithm DP2 is replaced by the recursive relation below, the modified Algorithm MDP2 solves AUX optimally. Modified DP2 (MDP2) V j1 jk = min V j1 −q1 jk −qk +

qi max maxpljl l = 1k Pljl m n i∈K l∈K k +ti $ Rlql jl l=1


620


+1− -f +g$q1 qk . all possible routes $(qi ≤ ji for

i = 1k( and 1 ≤ q1 +···+qk ≤ c Observe that in DP2 the departure time l∈K Pljl of k the shipment containing the jobs from l=1 Rl ql jl is replaced by the maximum of (i) and (ii) in MDP2. In the following heuristic, an LPT (largest-processing-time-first) order of a given set of jobs denotes that the jobs are sequenced in a nonincreasing order of their processing times pj . Also, we use the first available machine rule (FAM), where the first unscheduled job is scheduled to the earliest available machine of the processing facility until all the jobs are scheduled. Heuristic H1 Step 1. Reindex the jobs in Ni from each customer i ∈ K as i1 ini so that pi1 ≤ · · · ≤ pini . Define sets Qih and Riuv as in DP3 for h = 1 ni , v = u ni , u = 1 c, and i ∈ K. Step 2. Solve the auxiliary problem AUX using Algorithm MDP2. Let S1 SL be the shipments and $1 $L be the corresponding routes obtained by MDP2. Step 3. Sort the jobs in Sq in LPT order for q = 1 L. Apply the FAM rule to the jobs in S1 SL in this order. Let C-q. be the completion time of the last job in Sq for q = 1 L. For q = 1 L, deliver the jobs in Sq in one shipment (the qth shipment) departing at time C-q. and taking route $q . Step 2 of H1 is the most computationally intensive. Because MDP2 has the same structure as that of the original DP2, the overall time complexity of Heuristic H1 is also Oc k nk kk−1 . Therefore, for problems P3 and P4 with a fixed number k of customers, Heuristic H1 takes polynomial time. Given a solution 1 for the auxiliary problem AUX, we denote the mean delivery time of jobs, distribution cost, and total cost by DAUX 1 , TAUX 1 , and ZAUX 1 , respectively. Similarly, given a solution 1 to problem P ∈ P3 P4, these three values are denoted by DP 1 , TP 1 , and ZP 1 , respectively. Denote an optimal solution of P by 1P∗ and the optimal objective value by ZP∗ . Then, ZAUX 1 = DAUX 1 + 1 − TAUX 1 ZP 1 = DP 1 + 1 − TP 1 and ZP∗ = ZP 1P∗ . Lemma 4. Let 1AUX be the solution found by MDP2. Then, ZAUX 1AUX ≤ ZP∗ . Proof. Given an optimal solution 1P∗ for P3 or P4, suppose that there are V shipments. Let B1 BV be the subsets of jobs and 41 4V the respective routes associated with the V shipments in this solution. Let Ev be the production completion time of the last job in Bv for v = 1 V . Without loss of generality,

we assume that the shipments B1 BV are ordered in nondecreasing order of Ev . Then,

(1) pj m Ev ≥ max maxpj j ∈ B1 ∪···∪Bv j∈B1 ∪···∪Bv

and the mean delivery time of the jobs is DP 1P∗ =

V 1 B E + N ∩ Bv ti 4v Bv n v=1 v v i∈K i

(2)

where ti 4v Bv is defined in DP2 as the time required to reach customer i in route 4v . The solution 1P∗ is feasible for the auxiliary problem AUX if we set the departure time of shipment Bv for v = 1 V to be

pj m ev = max maxpj j ∈ B1 ∪ · · · ∪ Bv j∈B1 ∪···∪Bv

instead of Ev . By (1), Ev ≥ ev for v = 1 V . By (2), the mean delivery time of jobs in 1P∗ for AUX satisfies DAUX 1P∗ =

V 1 Bv ev + Ni ∩ Bv ti 4v Bv n v=1 i∈K

≤ DP 1P∗

(3)

Clearly, solution 1P∗ incurs the same transportation cost for P3, P4, and AUX, i.e., TP 1P∗ = TAUX 1P∗ . Together with (3), this implies that ZP∗ ≥ ZAUX 1P∗

(4)

Next, we show that the solution 1AUX generated in Step 2 for the auxiliary problem is optimal for that problem. Therefore, ZAUX 1P∗ ≥ ZAUX 1AUX , which, along with (4), establishes the lemma. In Step 1 of H1, jobs from each customer are sorted in SPT order, so the solution 1AUX satisfies the following customer-SPT property: The processing time of any job delivered in an earlier shipment is no more than the processing time of any job of the same customer delivered in a later shipment. Because the search space of this algorithm includes all the solutions satisfying the customer-SPT property, and the recursive relation compares all possible routes for a shipment, the solution 1AUX is optimal among all the feasible solutions that satisfy the customer-SPT property. To show that 1AUX is optimal for the auxiliary problem among all the feasible solutions, we only need to show that there exists an optimal solution to the auxiliary problem that satisfies the customerSPT property. To show this, consider a solution to the auxiliary problem that violates the customerSPT property. Suppose that in this solution there are U shipments B1 BU with corresponding routes 41 4U , and jobs j and l satisfying the following: j, l ∈ Ni , j ∈ Bx , l ∈ By , pj > pl , and x < y. If we exchange


621


shipments for these two jobs (i.e., deliver job j in By and job l in Bx , the departure time of each shipment will not increase, and the routes 4x and 4y are still feasible for Bx and By because jobs j and l belong to the same customer so no additional customer needs to be visited to deliver Bx or By . This means that the mean delivery time of jobs will not increase after the shipments of jobs j and l are exchanged. Clearly, the total transportation cost will remain the same. Hence, exchanging jobs j and l will not increase the total objective value. We can repeat this until the final solution satisfies the customer-SPT property. This shows that there exists an optimal solution to the auxiliary problem that satisfies this property. Theorem 4. Let 1P be the solution produced by H1 for P ∈ P3 P4 with 0 < < 1. Then, ZP 1P /ZP∗ ≤ 2 − 1/m, i.e., the worst-case performance ratio of H1 for P3 and P4 with 0 < < 1 is bounded by 2 − 1/m. Proof. The solution 1P constructed for P3 or P4 in Step 3 of Heuristic H1 has the same shipments S1 SL and routes $1 $L as but different shipment departure times than the solution 1AUX for the auxiliary problem. Therefore, TP 1P = TAUX 1AUX

(5)

The mean delivery time of jobs in 1AUX for the auxiliary problem is DAUX 1AUX =

L

1 Sl max max pj pj m j∈S1 ∪···∪Sl n l=1 j∈S1 ∪···∪Sl + Ni ∩Sl ti $l Sl (6) i∈K

and the mean delivery time of the jobs in 1P for P3 or P4 is L 1 DP 1P = S C + N ∩ Sl ti $l Sl (7) n l=1 l -l. i∈K i where C-l. is the production completion time of the jobs in shipment Sl for l = 1 L. For l = 1 L, let vl be the job that is completed last among all the jobs in S1 ∪ · · · ∪ Sl in Step 3 of H1. Due to the FAM rule in Step 3, we have 1 C-l. ≤ pj − pvl + pvl m j∈S1 ∪···∪Sl =

1 m−1 p p + m j∈S1 ∪···∪Sl j m vl

≤

1 m−1 pj + max p m j∈S1 ∪···∪Sl m j∈S1 ∪···∪Sl j

for l = 1 L (8)

By (6), (7), and (8), we have L 1 1 m−1 S max p p + DP 1P ≤ n l=1 l m j∈S1 ∪···∪Sl j m j∈S1 ∪···∪Sl j + Ni ∩ Sl ti $l Sl L

i∈K

1 1 S p + Ni ∩Sl ti $l Sl = n l=1 l m j∈S1 ∪···∪Sl j i∈K

L m−1 1 + S max p m n l=1 l j∈S1 ∪···∪Sl j m−1 ≤ DAUX 1AUX + DAUX 1AUX m 1 = 2− DAUX 1AUX m

(9)

This, together with (5), implies that ZP 1P = DP 1P + 1 − TP 1P ≤ 2 − 1/m -DAUX 1AUX + 1 − TAUX 1AUX . = 2 − 1/m ZAUX 1AUX Because ZAUX 1AUX ≤ ZP∗ by Lemma 4, we have ZP 1P ≤ 2 − 1/m ZP∗ . Next, we show that Heuristic H1 is asymptotically optimal under some mild conditions. Theorem 5. If the processing time pj of jobs j ∈ N follows a distribution over a finite -X Y ., where 0 < X < Y < and m is fixed, then the schedule 1p generated by Heuristic H1 is asymptotically optimal for problems P3 and P4 with 0 < < 1, as n approaches , i.e., ZP 1P − ZP∗ = 0 n→ ZP∗ lim

Proof. By (6), we have DAUX 1AUX ≥

L 1 Sl pj /m n l=1 j∈S1 ∪···∪Sl

l L X Sl Si nm l=1 i=1 L 2 l−1

L X = Sl S − Si nm l=1 l i=1 l=1 2 2

L L X 1 ≥ S − S nm l=1 l 2 l=1 l L 2 X X 2 nX = n = (10) Sl = 2nm l=1 2nm 2m ≥


622


Furthermore, by (6) and (9), we have DP 1P − DAUX 1AUX ≤ ≤

L m−1 1 Sl max pj j∈S1 ∪···∪Sl m n l=1

m−1 Y m

which, together with (10), implies lim

n→

DP 1P − DAUX 1AUX = 0 DAUX 1AUX

(11)

Because ZAUX 1AUX = DAUX 1AUX +1− ·TAUX 1AUX and ZP 1P = DP 1P +1− TP 1P , by (5) and (11), we have lim

n→

ZP 1P − ZAUX 1AUX = 0 ZAUX 1AUX

By Lemma 4 and the fact that ZP 1P ≥ ZP∗ , we have ZP 1P − ZP∗ = 0 n→ ZP∗ lim

This proves the theorem. In Theorem 5, we do not make any assumption on the number of customers k, the travel times tij , or the travel costs cij . Therefore, the result holds for any distributions of these parameters. Next, we conduct a computational test to evaluate the performance of Heuristic H1 for problems P3 and P4. Test problems are randomly generated as follows: (a) Weighting parameter in the objective function ∈ 02 05 08. (b) Number of jobs n ∈ 20 40 80 160, number of machines m ∈ 2 4 8, number of customers k ∈ 1 2 3 4 5 (the case with k = 1 corresponds to P3, and the cases with k ≥ 2 corresponds to P4), and capacity of each shipment c ∈ 4 8 12. Table 1

Computational Results for Heuristic H1 k =1

n 20

40

80

160

(c) Job processing times pj are drawn from the discrete uniform distribution U -1 100.. (d) The processing facility and the customers are located in a Euclidean square with length and width both equal to 9 ∈ 100 200 400. The processing facility is located at the center of the square, while the customers are independently and uniformly located within the square. The travel time tij is defined as the integer part of the Euclidean distance from location i to location j. The three values of 9 chosen here result in the average travel times between the processing facility and the customers that are less than, about equal to, or greater than the average processing time of a job, respectively. (e) The fixed transportation cost f is an integer from U -504 2504., and travel costs cij are integers from U -08tij 4 12tij 4., where 4 = 25n/m + 9 / -159 + 150 n/c. is a scaling parameter used to ensure that the two measures in the objective function, Dmean and T , are close to each other in the solution to the problem with = 05 generated by the heuristic so that these two measures are comparable. On average, Dmean should be close to 25n/m + 9, while T should be close to 159 + 150 n/c if 4 = 1 and all the shipments are full. Therefore, after scaling f and cij by the factor 4 Dmean is fairly close to T . We report both average and maximum relative gaps on test problems, where the relative gap on a test problem is defined as -ZP 1P − ZAUX 1AUX ./ ZAUX 1AUX , where 1P is the solution generated by Heuristic H1 and 1AUX the solution generated in Step 2 of H1. Table 1 shows the test results. For each combination of k n m reported in the table, we test 10 problems for each of the 27 combinations of the other parameters c, , and 9. Each entry in the columns “Avg gap” (“Max gap”) is the average (maximum) relative gap over the 270 random test problems for the corresponding k n m combination. The

k =2

k =3

k =4

k =5

m

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

2 4 8 2 4 8 2 4 8 2 4 8

087 214 347 049 093 303 008 035 206 005 019 116

408 940 1517 243 595 1223 113 188 653 020 087 365

101 239 357 049 147 342 024 062 233 010 029 135

455 965 1308 240 649 1161 114 320 702 057 146 417

087 224 361 051 150 342 029 081 260 011 035 143

365 887 1348 199 543 1051 111 290 718 057 156 382

092 255 364 055 157 335 026 080 261

383 880 1520 212 491 1191 115 284 685

079 254 393 056 159 344

389 856 1376 227 509 954


623


running time of the heuristic varies largely with the number of customers k in the problem. Each of the test problems with k ∈ 1 2 3, k = 4, and k = 5 is solved within 20, 60, and 150 CPU seconds, respectively. It typically takes significantly more CPU time (varying from 500 to a few thousand seconds) to solve a problem with k = 4 and n = 160, or k = 5 and n ∈ 80 160. Hence, no results are reported on these combinations of k and n. These results demonstrate that the heuristic is capable of generating near-optimal solutions within a reasonable amount of CPU time for large problems with k ∈ 1 2 3 and medium problems with k ∈ 4 5. They also show that the average relative gaps decrease with the number of jobs n and are very close to 0 when n ≥ 80. This computationally verifies the asymptotic optimality of H1.

4.

Solving Problems P5, P6, P7, and P8

In this section, we consider problems P5, P6, P7, and P8, where the customer service level is measured by the maximum delivery time Dmax . As in §3, we propose efficient algorithms for finding optimal solutions or show that they are NP-hard, and for two NP-hard problems, P7 and P8 with 0 < < 1, we propose a heuristic and analyze its worst-case and asymptotic performance. 4.1. Optimal Algorithms We treat the case with = 0 or 1 separately from the case with 0 < < 1. We first consider problem P5 with 0 < < 1. Because there is only one customer in P5, the maximum delivery time in an optimal schedule for this problem is simply Dmax = Cmax + t01 , where Cmax is the completion time of the last job in the production part. Hence, any production schedule without idle time is optimal. Also, because each shipment incurs the same distribution cost f + t01 + t10 , the optimal distribution schedule should have a minimum number of shipments, i.e., n/c shipments. Therefore, Algorithm A1 proposed in §3.1 can be applied. Remark 3. Algorithm A1 solves P5 with 0 < < 1 to optimality in On time. The total cost of the schedule generated by Al gorithm A1 is j∈N pj + t01 + 1 − n/cf + t01 + t10 . Next, we consider problem P6 with 0 < < 1. An extension of DP2 described in §3.1 can solve this problem. For ease of presentation, we use the same notation defined right before DP2 in §3.1. By Lemmas 2 and 3, the departure time of a shipment must be one of the times in the set i∈K Pihi 0 ≤ hi ≤ ni i ∈ K. The number of such times is bounded by nk . If the jobs in a shipment belong to u different customers, then there are u! possible routes for the shipment.

Summing over possible values of u, there are no more than 1! + · · · + k! ≤ k + 1 ! possible routes. Because the delivery time of each job in a given shipment is determined by the route and departure time of the shipment, there are at most k +1 !nk possible delivery times Dj for j ∈ N . Let : denote the set of all possible times Dj . Algorithm DP3 For every trial value of Dmax ∈ : , the following dynamic program finds a schedule with minimum total distribution cost T . Then, the schedule that minimizes Dmax + 1 − T is optimal to problem P6. Define value function V j1 jk = minimum total distribution cost of a schedule for the j1 + · · · + jk jobs from ki=1 Qi ji , given that all the jobs are delivered before or at Dmax . In addition, define g$ q1 qk as in DP2, and define tmax $ X to be the time required to reach the last customer in route $, which covers all the customers of the jobs in the set X. Initial condition: V 0 0 = 0. Recursive relation: For ji = 0 ni ; i = 1 2 k: V j1 jk = min V j1 − q1 jk − qk + 1 − f + g$ q1 qk all possible k Pljl + tmax $ Ri qi ji routes $ with l∈K

≤ Dmax ( qi ≤ ji for

i=1

i = 1 k( and 1 ≤ q1 + · · · + qk ≤ c Optimal solution value: V n1 nk . Theorem 6. Algorithm DP3 finds an optimal schedule for problem P6 with 0 < < 1 in Oc k n2k kk−1 time. Proof. The optimality of the dynamic programming algorithm with a given value of Dmax can be proved using similar arguments to the proof of Theorem 2 for DP2. By enumerating all possible values of Dmax , the overall Algorithm DP3 finds a schedule with the minimum total cost. Similar to DP2, the time required by DP3 with a given value of Dmax is bounded by Oc k nk kk−1 . Because there are at most k + 1 !nk possible values of Dmax and k!/kk ≤ 2/k2 when k ≥ 2, the overall algorithm requires no more than Oc k n2k kk−1 time. Theorem 6 implies that if the number of customers k is fixed, problem P6 with 0 < < 1 is solvable in polynomial time. However, if k is arbitrary, P6 with 0 < < 1 is strongly NP-hard because a special case of the problem reduces to the strongly NP-hard TSP problem. For P6, because Dmax ≥ j∈N pj + tmin ,where tmin = mint0i i ∈ K, the set ; = x ∈ : x ≥ j∈N pj + tmin contains all possible values of Dmax . Because Dmax ≤

624


j∈N pj +tmax , where tmax is the maximum time needed to reach a customer from the processing facility, there are at most mink + 1 !nk tmax − tmin different values in ;. We can speed up DP3 by minimizing the number of different values of Dmax in ; to be tried. We can sort all the Dmax values in nonincreasing order and try them in this order. Suppose that DP3 is run for Dmax = x. If in the resulting optimal solution the actual Dmax = y < x, then all the values in ; between x and y are disregarded. Although the worst-case time complexity of DP3 is much higher than that of DP2, our computational test shows that the actual running time of DP3 is only slightly higher than that of DP2. Problems P7 and P8 with 0 < < 1 are NP-hard because a special case of these problems reduces to the NP-hard parallel-machine scheduling problem Pm Cmax (Garey and Johnson 1979). Finally, we consider problems P5, P6, P7, and P8 with = 0 or = 1. When = 0, these problems are the same as P1, P2, P3, and P4 with = 0, respectively. Therefore, the corresponding results of §3.1 apply. When = 1, the objective is to minimize Dmax . In an optimal solution, each job is delivered in a separate shipment immediately after its processing. Therefore, Dmax = maxi∈K maxj∈Ni Cj + t0i . The following results are straightforward and hence stated without a proof. Remark 4. For P5 with = 1, any production schedule is optimal and Dmax = j∈N pj + t01 Remark 5. For P6 with = 1, it is optimal to schedule jobs from the same customer consecutively and jobs from different customers in the nonincreasing order of t0i . When tij = 0 for i, j ∈ 0 k, problems P7 and P8 with = 1 reduce to the NP-hard problem Pm Cmax , and hence P7 and P8 with = 1 are NP-hard.

4.2. Heuristic and Performance Analysis In this section, we propose a heuristic for problem P8 with 0 < < 1, which is then modified to solve P7 with 0 < < 1. Given an instance of problem P8, we define an auxiliary problem AUX as follows: AUX is the same as P8 except that there is only one machine in the processing facility, i.e., m = 1, and the processing time of each job j ∈ N is given as qj = pj /m, where pj is the processing time of job j in P8. Evidently, AUX is an instance of P6 and hence is solved by DP3. The heuristic first solves AUX and then transforms the resulting solution to a feasible solution of P8. Heuristic H2 Step 1. Reindex the jobs in Ni from each customer i ∈ K as i1 ini such that pi1 ≤ · · · ≤ pini . Define sets Qih Riuv , and : as in DP3, except that pj is replaced by qj for j ∈ N . Step 2. Solve problem AUX by DP3. Let S1 SL be the resulting shipments and $1 $L the corresponding routes.


Step 3. Apply the FAM rule to jobs in S1 SL in this order. For l = 1 L, deliver the jobs in Sl in one shipment departing at time C-l. and taking route $l , where C-l. is the completion time of the last job in Sl . The time complexity of H2 is dominated by Step 2 and is the same as in DP3, i.e., Oc k n2k kk−1 . For ease of presentation, we use the following notation in the remainder of this section. Given a solution 1 for problem P, we denote the maximum delivery time of jobs, distribution, and total cost by DP 1 TP 1 , and ZP 1 , respectively. Denote an opti∗ mal solution of P8 by 1P8 and the optimal objective ∗ value of P8 by ZP8 . By this notation, ZP 1 = DP 1 + ∗ ∗ 1 − TP 1 , and ZP8 = ZP8 1P8 . Lemma 5. Let 1AUX be the solution found by DP3 ∗ when applied to AUX . Then, ZAUX 1AUX ≤ ZP8 . ∗ Proof. Given an optimal solution 1P8 for P8, suppose that in this solution there are V shipments B1 BV with respective routes 41 4V . Let Ev be the production completion time of Bv for v = 1 V . Without loss of generality, we assume that E1 ≤ · · · ≤ EV . Because all jobs in B1 ∪ · · · ∪ BV are completed by time Ev , we have

Ev ≥

1 p m j∈B1 ∪···∪Bv j

(12)

∗ The maximum delivery time of the jobs in 1P8 is ∗ DP8 1P8 = max Ev + tmax 4v Bv v=1V

(13)

where tmax 4v Bv is defined in DP3 to be the time required to reach the last customer in route 4v . ∗ The solution 1P8 is feasible to the auxiliary problem AUX if we schedule the jobs in the order of the shipments B1 BV , with jobs in the same shipment scheduled in an arbitrary order, and set the departure time of shipment Bv as ev =

1 p m j∈B1 ∪···∪Bv j

instead of Ev for v = 1 V . From (12), Ev ≥ ev for v = 1 V . From (13), the maximum delivery time ∗ of the jobs in 1P8 for AUX satisfies ∗ ∗ DAUX 1P8 = max ev + tmax 4v Bv ≤ DP8 1P8 (14) v=1V

∗ incurs the same transportation Clearly, solution 1P8 ∗ ∗ cost in both AUX and P8, i.e., TP8 1P8 = TAUX 1P8 . This, together with (14), implies that ∗ ∗ ZP8 ≥ ZAUX 1P8

(15)

Because the solution 1AUX produced by H2 is opti∗ mal for AUX , ZAUX 1P8 ≥ ZAUX 1AUX , which, along with (15), establishes the lemma. Theorem 7. Let 1 be the solution produced by H2 for ∗ problem P8 with 0 < < 1. Then, ZP8 1 /ZP8 ≤ 2 − 1/m,


i.e., the worst-case performance ratio of H2 is bounded by 2 − 1/m. Proof. The solution 1 obtained by H2 has the same shipments S1 SL and routes $1 $L as but different shipment departure times from the solution 1AUX for problem AUX generated in Step 2 of H2. Therefore, TP8 1 = TAUX 1AUX (16) The maximum delivery time of the jobs in 1AUX for AUX is

1 DAUX 1AUX = max pj + tmax $l Sl (17) l=1L m j∈B1 ∪···∪Bl and the maximum delivery time of the jobs in 1 for P8 is (18) DP8 1 = max C-l. + tmax $l Sl l=1L

where C-l. is the production completion time of the jobs in shipment Sl in solution 1 for l = 1 L. Due to Step 3 of H2, C-l. ≤

1 m−1 p + max p m j∈S1 ∪···∪Sl j m j∈S1 ∪···∪Sl j

for l = 1 L (19) From (17), (18), and (19), we have 1 m−1 DP8 1 ≤ max max p pj + l=1L m m j∈S1 ∪···∪Sl j j∈S1 ∪···∪Sl

+ tmax $l Sl

1 ≤ max pj + tmax $l Sl l=1L m j∈S1 ∪···∪Sl +

m−1 max p m j∈N j

m−1 max p m j∈N j m−1 ∗ ≤ DAUX 1AUX + (20) DP8 1P8 m This, along with (16), implies that ZP8 1 ≤ ∗ . Because ZAUX 1AUX ≤ ZAUX 1AUX + 1 − 1/m ZP8 ∗ ∗ . ZP8 (by Lemma 5), we have ZP8 1 ≤ 2 − 1/m ZP8 By a proof similar to that of Theorem 5, it can be shown that H2 is asymptotically optimal under some mild assumptions on job processing times. Thus, we state the result without a proof. ≤ DAUX 1AUX +

Theorem 8. If the processing time pj of each job j ∈ N follows a distribution over a finite interval -X Y . for 0 < X < Y < , and m is fixed, then the schedule 1 generated by Heuristic H2 is asymptotically optimal for problem P8 with 0 < < 1 as n approaches , i.e., ∗ ZP8 1 − ZP8 = 0 ∗ n→ ZP8

lim

625

Note that the asymptotic optimality of Heuristic H2 is valid for any distribution of the number of customers k, travel times tij , and travel costs cij . Heuristic H2 can be slightly modified to solve problem P7 with 0 < < 1 as follows. Algorithm A1 is used instead of DP3 in Step 2. Evidently, the auxiliary problem is of type P5 instead of P6. Hence, the shipments S1 SL follow the same route. The complexity of this modified heuristic is On as in A1. The proofs of Lemma 5 and Theorems 7 and 8 (when P6 and P8 are replaced by P5 and P7, respectively) hold for this modified heuristic for P7 with 0 < < 1. To evaluate the performance of Heuristic H2 for problems P7 and P8, we conduct an experiment in which test problems are generated in the same way as in the experiment for Heuristic H1 in §3.2, except that the scaling parameter 4 is defined as 4 = 50n/m + 9 / -159 + 150 n/c.. On average, Dmax should be close to 50n/m + 9, while T should be close to 159 + 150 n/c if 4 = 1 and all the shipments are full. Therefore, after scaling all the parameters f and cij by the factor 4, Dmax should be fairly close to T . In Table 2 we report the average and maximum relative gap between the objective value of the solution generated by H2 and the lower bound ZAUX 1AUX based on 270 randomly generated test problems (10 for each of the 27 combinations of c, , and 9). Each of the test problems with k ∈ 1 2 3, k = 4, and k = 5 is solved within 30, 80, and 300 CPU seconds, respectively. Problems with k = 4, n = 160, and problems with k = 5, n ∈ 80 160 require significantly more computational time; hence, no results are reported for these problems. These results show that H2 is capable of generating near-optimal solutions within a reasonable CPU time for large problems with k ∈ 1 2 3 and medium problems with k ∈ 4 5. Moreover, the relative gap decreases with the number of jobs n and is very close to 0 when n ≥ 80. This computationally verifies the asymptotic optimality of H2 (Theorem 8).

5.

Value of Integration

The problems considered in §§3 and 4 integrate production and distribution scheduling. However, in practice, production and distribution decisions are often made separately and sequentially. Most production scheduling models only consider job processing and seek to optimize a time-related performance measure that is usually a function of job completion times. On the other hand, most vehicle-routing models assume that jobs to be delivered have already been processed and are only concerned with minimizing the total distribution cost. In this section, we evaluate the value of productiondistribution integration by comparing the integrated approach proposed in this paper with a typical sequential approach that addresses production and distribution sequentially. In a typical sequential


626


Table 2

Computational Results for Heuristic H2 k =1

n 20 40 80 160

k =2

k =3

k =4

k =5

m

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

Avg gap (%)

Max gap (%)

2 4 8 2 4 8 2 4 8 2 4 8

018 076 678 005 020 291 002 008 162 001 002 087

100 322 2465 028 092 890 009 032 507 002 008 228

037 180 724 019 066 343 003 015 171 001 005 094

236 929 2881 162 493 1429 013 065 618 009 070 319

045 194 778 025 089 400 012 039 213 002 006 099

275 1130 3293 267 540 1586 153 380 954 007 038 343

057 249 781 028 100 393 014 043 178

525 1624 2997 192 660 1346 101 213 786

045 293 861 038 116 450

309 1401 3147 166 443 1210

approach (SA), the production part assumes that each job j ∈ Nu once completed is delivered to its customer immediately, i.e., Dj = Cj + t0u , and optimizes the customer service level (i.e., minimizes Dmean or Dmax ) based on that assumption without considering possible delivery consolidation of jobs. The distribution part in this approach minimizes the total transportation cost T given the production schedule fixed by the production part. 5.1. Sequential Approach for P1, P2, P3, and P4 If SA is used to solve problems P1, P2, P3, and P4, because it is assumed in the production part that Dj = Cj + t0u for j ∈ Nu , minimizing Dmean is equivalent to minimizing j∈N Cj . Then, scheduling using the FAM rule on the SPT order of jobs is optimal. Given this SPT production schedule, the distribution part is a generalized vehicle-routing problem where jobs are ready for delivery at the times specified by the SPT schedule. This routing problem can be solved by DP2 in §3.1, after the value function V j1 jk is redefined as the minimum total transportation cost for delivering the j1 + · · · + jk jobs from ki=1 Qi ji and the parameter is replaced by 0. The departure time of each shipment is equal to the completion time of the last job in this shipment given in the SPT production schedule. The objective value of the entire problem Dmean + 1 − T can be calculated accordingly. 5.2. Sequential Approach for P5, P6, P7, and P8 If SA is used to solve problems P5 and P6 where there is only one machine, any production schedule without idle time minimizes Dmax = maxCj + t0u j ∈ Nu u ∈ K and hence is optimal for the production part. However, for problems P7 and P8 where there are multiple machines, minimizing Dmax is at least as hard as the NP-hard problem Pm Cmax . Therefore, it is not practical to solve this problem to optimality. We propose the following heuristic for this problem.

Heuristic SA-H Step 0. Reindex customers such that t01 ≥ t02 ≥ · · · ≥ t0k . Let i = 1. Step 1. Sort the jobs from customer i in LPT order and apply the FAM rule to schedule these jobs on the machines. Set i = i + 1, and repeat Step 1 until all customers are covered. A simple interchange argument shows that there exists an optimal schedule for the production part of SA, where the jobs of a customer i are scheduled earlier than jobs of another customer h if t0i > t0h . The schedule generated by SA-H satisfies this property. It can be shown that the optimal objective value of the problem (i.e., Dmax is greater than or equal to maxRi /m + t0i i ∈ K, where Ri is the total processing time of the jobs from customers 1 i. We tested SA-H on the exact same test problems used in §4.2 and found that the relative gap between the value of Dmax generated by Heuristic SA-H and the lower bound maxRi /m + t0i i ∈ K are on average no more than 0.5%, and in many cases are below 0.1%. Hence, we do not tabulate detailed results of SA-H other than the conclusion that SA-H produces nearly optimal schedules for the production part of the sequential approach for P7 and P8. Given an arbitrary production schedule for P5 and P6, or the schedule obtained by SA-H for P7 and P8, the distribution part of the sequential approach involves exactly the same problems as in the case of P1, P2, P3, and P4, respectively. Algorithm DP2 of §3.1 can be applied directly as explained in §5.1. Thus, the objective value of the entire problem Dmax + 1 − T can be calculated accordingly. 5.3. Comparisons and Insights In this section, we conduct a computational test to evaluate the typical benefits that can be brought by the integrated approach proposed in this paper relative to SA. For the integrated approach for problems P1, P2, P3, and P4, which seek to minimize the objective function Dmean + 1 − T , we use


627


Algorithms DP1 and DP2, and Heuristic H1, respectively. Applying these algorithms, either the optimal objective value (in the case of P1 and P2) or an upper bound of the optimal objective value (in the case of P3 and P4) can be found. Denote the objective value found by these algorithms by ZINT . On the other hand, these problems can also be solved by the SA described in §5.1. Denote the objective value found by SA by ZSEQ . Then, the relative improvement (i.e., relative cost reduction) from SA to the integrated approach for the objective function Dmean + 1 − T is defined as ZSEQ − ZINT /ZSEQ . Similarly, for problems P5, P6, P7, and P8, which seek to minimize the objective function Dmax + 1 − T , we use Algorithms A1 and DP3, and Heuristic H2, respectively, as the integrated approach. Denote the objective value found by these algorithms by ZINT , and denote the objective value found by SA described in §5.2 by ZSEQ . We can similarly calculate the relative improvement from ZSEQ to ZINT . In our computational experiment, test instances for problems with the objective function Dmean +1− T are generated exactly in the same way as in the experiment for Heuristic H1 in §3.2, except that the same three cases of n ∈ 20 40 60 are used for different cases of k ∈ 1 2 3 4 5. Similarly, test instances for problems with the objective function Dmax + 1 − T are generated exactly in the same way as in the experiment for Heuristic H2 in §4.2, except that the same three cases of n ∈ 20 40 60 are used for different cases of k ∈ 1 2 3 4 5. Our test shows that the relative improvement from SA to the integrated approach is primarily influenced by the number of customers k, capacity of the shipment c, and the weighting parameter . Therefore, we report results based on k c . Table 3 shows the average relative improvement from SA to the integrated approach. Each entry is the average taken over 270 test problems for the corresponding k c combination, 10 for each of the Table 3

27 combinations of n m 9 . From these results, we can make the following observations: (1) For problems with the objective function Dmean + 1 − T , the relative improvement from SA to the integrated approach is more than 5% for most of the test problems with k ≥ 2, and more than 10% for most of the test problems with k ≥ 2 and ≥ 05. A 5% increase in customer service level or a 5% reduction in total cost often represents a significant performance improvement in practice, thus justifying the merits of integration in our model. (2) For problems with the objective function Dmean + 1 − T , the relative improvement due to integration increases with the number of customers k, shipment capacity c, and the weighting parameter . This can be explained as follows. As the number of customers k or/and the shipment size c increase, there are more opportunities to consolidate jobs from different customers to joint deliveries. The integrated approach fully exploits these opportunities to achieve an optimal balance between Dmean and T . On the other hand, SA favors job consolidation merely from the distribution perspective. As a result, the schedule generated by SA often provides poor service. Therefore, the advantage of the integrated approach over SA becomes more significant when there are more consolidation possibilities. Also, because SA generates a schedule that minimizes the total distribution cost T , the performance of this schedule for the overall objective function Dmean + 1 − T deteriorates as increases. (3) For problems with the objective function Dmax + 1 − T , the relative improvement is negligible for problems with k = 1 but more than 5% for most of the test problems with k ≥ 3 and ≥ 05. In addition, the relative improvement increases with the weighting parameter but does not seem to be influenced much by the shipment capacity c. In terms of the number of customers k, the relative improvement increases when

Average Relative Improvement from Sequential to Integrated Approaches Dmean + 1 − T

Dmax + 1 − T

k

c

= 02 (%)

= 05 (%)

= 08 (%)

k

c

= 02 (%)

= 05 (%)

= 08 (%)

1

4 8 12 4 8 12 4 8 12 4 8 12 4 8 12

080 092 095 237 346 445 314 531 640 470 707 821 525 757 961

202 234 241 607 888 1105 797 1371 1632 1193 1811 2103 1371 2018 2637

323 369 405 1002 1516 1931 1331 2389 3013 1974 3187 3837 2357 3450 4396

1

4 8 12 4 8 12 4 8 12 4 8 12 4 8 12

000 0001 0008 182 177 137 289 216 187 264 258 157 313 261 160

000 0003 002 458 477 196 741 578 545 681 698 420 808 504 560

000 0004 003 742 792 364 1214 996 1100 1145 1284 839 888 1202 1129

2 3 4 5

2 3 4 5


628


k increases from 1 to 2 and from 2 to 3. However, the relative improvement does not vary much when k increases from 3 to 4 or 5. Although we did not test problems with k ≥ 6 because of the excessive run time required by Heuristic H2, the trend indicates that the overall relative improvement for problems with k ≥ 6 will probably be similar to that for problems with k ∈ 3 4 5. We conclude that the relative improvement is significant for problems with the objective function Dmax + 1 − T when k ≥ 3 and ≥ 05. (4) The relative improvement for the objective function Dmax + 1 − T is less pronounced. This is because the delivery schedule of every shipment has a direct impact on Dmean and T but not necessarily on Dmax . This means that scheduling jobs optimally generally makes a bigger difference if one tries to minimize Dmean + 1 − T instead of Dmax + 1 − T .

6.

Conclusions

We have proposed an integrated scheduling model of production and distribution operations and studied eight related problems. For each of the problems, we either proposed a polynomial time exact algorithm or proved its intractability. For the NP-hard problems, we proposed heuristics, analyzed their worst-case performance, and proved their asymptotic optimality under mild conditions on job processing times. We conducted computational experiments that demonstrate that the heuristics are capable of generating near-optimal solutions within a reasonable CPU time. We also investigated the value of integration by comparing it to a typical sequential approach. Our computational experiment shows that in most cases a significant improvement can be achieved by integration when the objective function is Dmean + 1 − T , and in some cases when the objective function is Dmax + 1 − T . The worst-case time complexity of our heuristics is exponential in number of customers. Future research is needed to answer the question of whether there exists a polynomial time heuristic with a constant worst-case performance ratio for our problems with an arbitrary number of customers. Future research may also consider applications where the firm fully owns the fleet of vehicles. Then, the number of concurrent shipments will be limited by the number of trucks available. Alternative machine configurations like flowshop facilities are of interest. Alternative production-distribution objectives are also desirable for capturing more diverse applications. Acknowledgments

The authors appreciate the constructive suggestions provided by three anonymous referees and the department editor, Dorit Hochbaum. The first author was supported in part by the National Science Foundation under grants DMI0196536 and DMI-0421637.

References Bramel, J., D. Simchi-Levi. 1997. Logic of Logistics: Theory, Algorithms, and Applications for Logistics Management. SpringerVerlag, New York. Chandra, P., M. L. Fisher. 1994. Coordination of production and distribution planning. Eur. J. Oper. Res. 72 503–517. Chen, Z.-L. 1996. Scheduling and common due date assignment with earliness-tardiness penalties and batch delivery costs. Eur. J. Oper. Res. 93 49–60. Chen, Z.-L. 2004. Integrated production and distribution operations: Taxonomy, models, and review. D. Simchi-Levi, D. Wu, Z.-J. Shen, eds. Handbook of Quantitative Supply Chain Analysis: Modeling in the E-Business Era. Kluwer Academic Publishers, Boston, MA. Cheng, T. C. E., V. S. Gordon, M. Y. Kovalyov. 1996. Single machine scheduling with batch deliveries. Eur. J. Oper. Res. 94 277–283. Cohen, M. A., H. L. Lee. 1988. Strategic analysis of integrated production-distribution systems: Models and methods. Oper. Res. 36 212–228. Conway, R. W., W. L. Maxwell, L. W. Miller. 1967. Theory of Scheduling. Addison-Wesley, Reading, MA. Feld, W. M. 2000. Lean Manufacturing: Tools, Techniques and How to Use Them. CRC Press, Boca Raton, FL. Fumero, F., C. Vercellis. 1999. Synchronized development of production, inventory, and distribution schedules. Transportation Sci. 33 330–340. Garey, M. R., D. S. Johnson. 1979. Computers and Intractability: A Guide to the Theory of NP-Completeness. W. H. Freeman, San Francisco, CA. Geismar, H. N., L. Lei, C. Sriskandarajah. 2003. The integrated production and transportation scheduling problem for a product with a short life span and non-instantaneous transportation time. Working paper, University of Texas at Dallas, Dallas, TX. Hahm, J., C. A. Yano. 1995. Economic lot and delivery scheduling problem: Powers of two policies. Transportation Sci. 29 222–241. Hall, N. G., C. N. Potts. 2003. Supply chain scheduling: Batching and delivery. Oper. Res. 51 566–584. Hall, L. A., D. B. Shmoys. 1992. Jackson’s rule for single-machine scheduling: Making a good heuristic better. Math. Oper. Res. 17 22–35. Hall, N. G., M. Lesaoana, C. N. Potts. 2001. Scheduling with fixed delivery dates. Oper. Res. 49 134–144. Herrmann, J., C.-Y. Lee. 1993. On scheduling to minimize earlinesstardiness and batch delivery costs with a common due date. Eur. J. Oper. Res. 70 272–288. Karp, R. M. 1972. Reducibility among combinatorial problems. R. E. Miller, J. W. Thatcher, eds. Complexity of Computer Computations. Plenum Press, New York, 85–103. Lee, C.-Y., Z.-L. Chen. 2001. Machine scheduling with transportation considerations. J. Scheduling 4 3–24. Li, C.-L., G. Vairaktarakis, C.-Y. Lee. 2005. Machine scheduling with deliveries to multiple customer locations. Eur. J. Oper. Res. 164 39–51. Pinedo, M. 2002. Scheduling: Theory, Algorithms, and Systems, 2nd ed. Prentice Hall, Upper Saddle River, NJ. Potts, C. N. 1980. Analysis of a heuristic for one machine sequencing with release dates and delivery times. Oper. Res. 28 1436–1441. Rajagopalan, S., A. Malhotra. 2001. Have U.S. manufacturing inventories really decreased? An empirical study. Manufacturing Service Oper. Management 3 14–24. Sarmiento, A. M., R. Nagi. 1999. A review of integrated analysis of production-distribution systems. IIE Trans. 31 1061–1074. Woeginger, G. J. 1994. Heuristics for parallel machine scheduling with delivery times. Acta Inform. 31 503–512. Woeginger, G. J. 1998. A polynomial-time approximation scheme for single-machine sequencing with delivery times and sequence-independent batch set-up times. J. Scheduling 1 79–87.